176 S. P. Langley on the Selective Absorption of Solar Energy. 



Example of the Mode of Observation. 



As an example of the first class of measures, let us consider 

 the observations made with the Hilger prism on June 22, 

 1882. The high-sun observation was made at b 15 m . The 

 sun's zenith-distance at this time was 17° 1CK; the air-mass* 

 was 1*047 time the mass overhead ; the height of the baro- 

 meter corresponding to the air-mass overhead was 7*39 de- 

 cimetres ; consequently the air-mass for a zenith-distance of 

 17° 10' was 7*39 x 1-047 = 7-74 decimetres. 



The sun's zenith-distance at 6 h 25 m (the time of the second 

 observation) was 79° 8' t; the height of the barometer was 

 the same as at noon; and the air-mass by the same formula 

 was 5*18 times that overhead, or 7*39x5*18 = 38*27 decime- 

 tres; so that the mass of air traversed in the second observation 

 exceeded that in the first by an amount capable of supporting 

 30*53 decimetres of mercury. 



The galvanometer-deflection obtained in the part of the 

 spectrum whose deviation is 44° 30' (a part which is near the 

 extreme lower limit of the present observations, far below the 

 visible red) was at noon 17, and in the afternoon 11. In the 

 violet, where the deviation is 50° 00', the corresponding de- 

 flections were 4*5 and 0*39. Let us take these two feeble 

 extreme rays as types with which to illustrate our process. 

 Considering first the infra-red ray, we have deflection at noon 

 = ^ = 17, deflection in afternoon =d 2 = l\, difference in mass 

 of air traversed = M 2 /3 2 — - MiA = 30*53 decimetres, which by 

 its absorption has produced the difference in the deflections. 

 t representing the amount of energy transmitted by a layer 

 of air equivalent to 1 decimetre of mercury, we find from the 

 formula 



f = *986 ; that is, a mass of air capable of supporting 1 deci- 

 metre of mercury in the barometer transmits 98*6 per cent, of 

 the energy of this particular kind of ray. This quantity t we 

 call the coefficient of transmission of the ray. 



Knowing now the amount of energy transmitted by one 

 such layer of air, we can find the amount transmitted by the 

 7*74 layers which intervened between the observer and the 

 sun at noon, namely *986 774 = *895. Only 89*5 per cent., 



* Computed from the formula M= tabular refraction 



cosapp. altitude, 

 t In general it is not advisable to make observations at so great a 

 zenith-distance as this. 



